(2) CaH2 (4) NH3
2. Hypothesis
3. Pollutant
4. Matter
b. carbon
c. sodium
d. iodine
The answer to your question is
d. idoine
(1) Group 1, Period 4 (3) Group 2, Period 3
(2) Group 2, Period 5 (4) Group 3, Period 4
Answer: Option (2) is the correct answer.
Explanation:
Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.
And, when we move across a period then there is a decrease in the ionization energy due to decrease in the size of atoms.
Whereas when we move down a group then there is an increase in ionization energy because of increase in size of the atoms.
Since, it is given that each atom has 2 valence electrons therefore, all the atoms belong to group 2.
And, as it has lower first ionization energy than calcium it means the atom is larger in size than calcium atom.
Hence, we can conclude that this element located in Group 2, Period 5 on the periodic table.
The element with two valence electrons and a first ionization energy lower than calcium would be found in Group 2, Period 5 on the Periodic Table.
The element you're referring to, with two valence electrons and a lower first ionization energy than calcium, would be located in Group 2, Period 5 on the Periodic Table.
Here's why: Group 2 elements, or alkaline earth metals, all have two valence electrons. Periods on the other hand, describe the energy level of an atom's outermost electrons, with higher periods corresponding to higher energy levels. Thus, an atom in a higher period would have a lower ionization energy (the energy required to remove an electron) than one in a lower period.
Therefore, because the first ionization energy is lower than calcium (which is in Group 2, Period 4) but still has two valence electrons, this element would be located one period below, in Group 2, Period 5.
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